3

The AMS style guide, p.118, says "If there is a long expression before the first verb, align succeeding verbs with a two-em quad indent from the left."

enter image description here

Right now I can achieve this with a & on the first line and &\hspace{2em} on all succeeding lines. Is there some way to achieve it with &'s alone or do I have to keep writing \hspace{2em}?

4
  • You could replace \hspace{2em} with \qquad...
    – Mico
    Jul 13 at 2:29
  • @Mico OK, thanks for the shortcut
    – Hasse1987
    Jul 13 at 4:42
  • 1
    The 2em "rule" is useful, but it's not a fundamental law of nature. Just settle on a spacing rule you can live with.
    – Mico
    Jul 13 at 17:59
  • @MadPhysicist - Thanks for noticing! I've posted a corrected version of the comment.
    – Mico
    Jul 13 at 17:59

2 Answers 2

5

You can use \MoveEqLeft from mathtools.

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\[
\begin{split}
\MoveEqLeft xxxxxxxxxxxxxxxxx \\
& = xxx + xxx \\
& = xx + xxxxxx.
\end{split}
\]
\end{document}

equation example

If you give it without the optional argument it is moved [2em]. But you can also say \MoveEqLeft[3] to move 3em.

3

For the multirow equation at hand, you could place the & alignment symbol in front of f(x) instead of in front of \abs[\bigg]{...}.

enter image description here

\documentclass{article} % or some other suitable document class

\usepackage{mathtools,amssymb}
\DeclarePairedDelimiter\abs\lvert\rvert
\DeclarePairedDelimiter\norm\lVert\rVert

\begin{document}

\begin{align*}
\abs[\bigg]{\int_{\mathbb{R}^n}
  &f(x)g_N(x)\,d\mu(x)} \\
  &\le \sum_{k=1}^{\infty} \abs{\lambda_k}\,\norm{a_k}_{L_p^q} 
       \biggl(\int_{\!S_k} \abs{{}\cdots{}}^{q'} d\mu(x)\biggr)^{\!1/q'} \\
  &\le \sum_{k=1}^{\infty} \abs{\lambda_k} \cdots
\end{align*}

% OP's version
\begin{align*}
&\abs[\bigg]{\int_{\mathbb{R}^n} f(x)g_N(x)\,d\mu(x)} \\
&\qquad\le \sum_{k=1}^{\infty} \abs{\lambda_k}\,\norm{a_k}_{L_p^q} 
       \biggl(\int_{\!S_k} \abs{{}\cdots{}}^{q'} d\mu(x)\biggr)^{\!1/q'} \\
&\qquad\le \sum_{k=1}^{\infty} \abs{\lambda_k} \cdots
\end{align*}

\end{document}
2
  • That's because you know everything up to the $f(x)$ to measure 2em?
    – Hasse1987
    Jul 13 at 4:34
  • @Hasse1987 - Happy coincidence, no more.
    – Mico
    Jul 13 at 5:28

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